How many arguments takes the foldr function of Haskell?

In this situation, I think it is best to look at the type signature of foldr:

foldr :: (a -> b -> b) -> b -> [a] -> b

and to match that to the expression we have (with added parenthesis for clarity):

(foldr step (\_ -> [[]]) xs) (\_ -> False)

The second argument of foldr is the same type as its result. In this case the second argument is a function. In this case, this means that the foldr expression with 3 arguments will be a function.

What you see to be the 4th argument of the foldr function could also be thought of as the 1st argument of the foldr result!

All functions in Haskell take just one argument. When we have a function with type a -> b -> c, it is just a shorter way to write a -> (b -> c), i.e. a function, which takes one argument and produces a function which takes another argument. See Currying for more information.

In this case, see the @sepp2k's answer. foldr produces a function and it needs another ("the 4th") argument.

In this case foldr is used to build up a function. (\_ -> False) is the argument to that function.